# Einstein summation convention: Del operator and dot product

Now, I am aware of the summation convention for the dot product

$$\mathbf{a} \cdot \mathbf{b} = a_i b_i$$

But I am unsure about how to represent $(\nabla \cdot \mathbf{a}) \mathbf{b}$ and $(\mathbf{a} \cdot \nabla) \mathbf{b}$, since in general, $$(\nabla \cdot \mathbf{a}) \mathbf{b} \ne (\mathbf{a} \cdot \nabla) \mathbf{b}$$

In either case, I come up with $\nabla_i a_i b_j$ = $a_i \nabla_i b_j$, since (from what I understand) the order of the terms ought not to matter. This is ambiguous, though, as it doesn't differ between the two. Can anyone hint me in the correct direction?

The order of the terms can matter. This is especially true of derivative operators and the objects on which they operate. For example,

\begin{align} (\nabla \cdot \vec a)\vec b&=(\partial_i a_i)b_j\\\\ &=b_j(\partial_i a_i)\\\\ &\ne a_i\partial_i(b_j) \,\,\text{(and certainly not}\,\,a_ib_j\partial_i)\\\\ &=(\vec a\cdot \nabla)\vec b \end{align}

I have actually found retaining unit vectors where applicable can help. In the previous example, I might have written

$$(\nabla \cdot \vec a)\vec b=(\partial_i a_i)(\hat x_jb_j)$$

For more complex expressions, this can be quite useful. For example, suppose we have $\vec A \times (\vec B\times \vec C)$. Then, I would write

\begin{align} \vec A \times (\vec B\times \vec C)&=\hat x_i A_i\times (\hat x_j B_j\times \hat x_kC_k)\\\\ &=(\hat x_i\times (\hat x_j\times \hat x_k))A_iB_jC_k\\\\ &=(\delta_{ik}\hat x_j-\delta_{ij}\hat x_k)A_iB_jC_k\\\\ &=(A_iC_i)\hat x_jB_j-(A_iB_i)\hat x_kC_k\\\\ &=(\vec A\cdot \vec C)\vec B-(\vec A\cdot \vec B)\vec C \end{align}

recovering the familiar vector triple product!

And in one last example, from THIS ANSWER I POSTED HERE

\begin{align} \nabla \times \nabla \times \vec A&=(\partial_i \hat x_i)\times (\partial_j \hat x_j)\times (\hat x_k A_k)\\\\ &=\hat x_i\times(\hat x_j\times \hat x_k)\partial_i\partial_j(A_k) \tag 1\\\\ &=\left(\delta_{ik}\hat x_j-\delta_{ij}\hat x_k\right)\partial_i\partial_j(A_k)\tag 2\\\\ &=\hat x_j\partial_j\partial_iA_i-\hat x_k\partial^2_i(A_k) \tag 3\\\\ &=(\hat x_j \partial_j)(\partial_i A_i)-\partial^2_i(\hat x_kA_k) \tag 4\\\\ &=\nabla \nabla \cdot \vec A-\nabla^2\vec A \tag5 \end{align}

In going from $(1)$ to $(2)$ we made use of the vector triple product. Note that $\delta_{ij}$ is the Kronecker Delta with $\delta_{ij}=1$ for $i=j$ and $0$ otherwise.

In going from $(2)$ to $(3)$, we used the sifting property of the Kronecker Delta.

In going from $(3)$ to $(4)$, we rearranged terms.

In going from $(4)$ to $(5)$, we recognized the terms of the final result in terms of their tensor representations.